# A doubt on Root space decomposition

Let $$H\subseteq T$$ be a maximal Toral subalgebra of a semsimple Lie algebra $$L.$$ Nowto obtain the so called root space decomposition one proceeds as follows. It is clear that $$ad_LH$$ consists of mutually commuting elements. My question is why $$L$$ can be written as $$L=\oplus_\alpha L_\alpha$$ where $$L_\alpha:\{x\in L:[hx]=\alpha(h)x, h\in H\}$$ where $$\alpha$$ ranges over $$H^*$$? I understand this is a generalization of the usual decomposition of diagonalizable matrix generalized to the case of commuting family of matrices. Bu how to prove this rigorously?

## 1 Answer

I'm assuming here we're talking about Lie Algebras over some an algebraically closed field $$\mathsf{k}$$, so that a maximal toral subalgebra is abelian. This result then relies on a more general lemma which goes as follows.

Let $$\mathfrak{h}$$ be a nilpotent Lie algebra with $$(V,\rho)$$ a finite dimensional representation of $$\mathfrak{h}$$. Then $$V$$ decomposes as a direct sum

$$V = \bigoplus_{\lambda \in \left( \mathfrak{h}/D\mathfrak{h}\right)^{*}} V_{\lambda},$$

where

$$V_{\lambda} = \left\{v \in V \mid \forall \ x \in \mathfrak{h}, \exists n \in \mathbb{N} : \left( \rho(x) - \lambda(x) \right)^n(v) =0 \right\}.$$

To prove this result, we proceed by induction on the dimension of $$V$$, with the case that the dimension of $$V$$ is $$1$$ being trivial.

Suppose now that the dimension of $$V$$ is greater than $$1$$, and suppose for the moment that $$V$$ admits a decomposition of the form $$V = U \oplus W$$ for some non-zero proper subrepresentations $$U,W$$, then by the inductive hypothesis then claim holds for $$U,W$$. But then by voting that $$V_{\lambda} = U_{\lambda} \oplus W_{\lambda}$$, we would get that the result holds for $$V$$ too.

Now consider $$x \in \mathfrak{h}$$, then $$\rho(x) : V \to V$$ is a linear map, and so $$V$$ admits a generalised eigenspace decomposition of the form $$V = \bigoplus V_{\lambda(x)}$$. It can be shown (this is one place where we need that $$\mathfrak{h}$$ is nilpotent) that each $$V_{\lambda(x)}$$ is in fact a subrepresentation of $$V$$, and so if it were the case that $$\rho(x)$$ had two distinct eigenvalues, then by the above discussion we would be done, and so we may reduce to the case that every $$x \in \mathfrak{h}$$ has precisely one eigenvalue $$\lambda(x)$$ say. Now Lie's theorem gives some $$0 \neq v \in V$$ and a Lie algebra homomorphism $$\mu : \mathfrak{h} \to \mathsf{k}$$ such that $$\rho(y)v = \mu(y)v$$ for all $$y \in \mathfrak{h}$$. But then $$\mu(x)$$ is an eigenvalue of $$\rho(x)$$, and so $$\mu(x) = \lambda(x)$$ for any $$x \in \mathfrak{h}$$, and so in particular the map $$x \to \lambda(x)$$ is in fact a member of $$(\mathfrak{h} / D\mathfrak{h} )^{*}$$. But what we have proven here is precisely that $$V = V_{\lambda}$$, and so we're done.

Now this is almost the result you want, but we have a bit more work to prove that the $$n$$s are all equal to $$1$$ in your case where $$\mathfrak{h} = H$$, $$V = L$$ and $$\rho = \operatorname{ad}_L$$.

One way to see this is to prove that in this case the non-zero $$L_{\alpha}$$ are all one-dimensional, and this can be proven by a dimension counting argument making use of the killing form, and the $$\mathfrak{sl}_{\alpha}$$-subalgebra of $$L$$.

See if you can conclude the proof from here.